D. Substring
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a graph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is 3. Your task is find a path whose value is the largest.
Input
The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.
The second line contains a string s with only lowercase English letters. The i-th character is the letter assigned to the i-th node.
Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between x and y. Also the graph can be not connected.
Output
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
Examples
Input
5 4
abaca
1 2
1 3
3 4
4 5
Output
3
Input
6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4
Output
-1
Input
10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7
Output
4
Note
In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter 'a' appears 3 times.
題目大意：給定一個有向圖，每個點有一個權值，用一個字母表示。一條路徑的權值就是這條路徑上出現次數最多的字母的個數。題目要求找出給定有向圖的最大權值。
當圖中存在環時，則權值可爲無窮大，此時無解。可用拓撲排序判環。
排除上面情況，有向圖必爲DAG。每次選擇入度爲零的點，保證每條路徑都從頭開始搜，DFS DP答案即可。搜過的點可直接合並答案，不必再搜下去，保證了複雜度O(m)。取最大值即爲答案。
代碼如下：
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
typedef struct node
{
int v,tol;
node(int vv = 0,int tt = 0): v(vv),tol (tt){}
bool operator < (const node & t) const
{
return tol > t.tol;
}
}node;
const int maxn = 3e5 + 5;
// 拓撲排序判環
// 找入度爲零的點記憶化搜索
vector<int> G[maxn];
int in[maxn],in_tsp[maxn];// 入度
int dp[maxn][26];// 記錄dp結果
bool vis[maxn];// 標記數組
bool TPS(int sum)
{
priority_queue<node> q;
for(int i = 0;i < sum; ++i)
{
q.push(node(i,in[i]));
in_tsp[i] = in[i];
}
node p;
while(!q.empty() && sum)
{
p = q.top();
q.pop();
int x = p.v;
if(p.tol > 0) return false;
if(vis[x]) continue;
vis[x] = true;
int len = G[x].size();
for(int i = 0;i < len; ++i)
{
int u = G[x][i];
if(!vis[u]) q.push(node(u,--in_tsp[u]));
}
--sum;
}
return true;
}
void DFS(int x)
{
int len = G[x].size();
int tp[26] = {0};
for(int i = 0;i < len; ++i)
{
int u = G[x][i];
if(!vis[u])
{
vis[u] = true;
DFS(u);
}
for(int j = 0;j < 26; ++j) tp[j] = max(tp[j],dp[u][j]);
}
for(int i = 0;i < 26; ++i) dp[x][i] += tp[i];
}
int main()
{
int x,y;
int n,m;
scanf("%d %d",&n,&m);
for(int i = 0;i < n; ++i)
{
vis[i] = false;
in[i] = 0;
for(int j = 0;j < 26; ++j) dp[i][j] = 0;
}
char p;
getchar();
for(int i = 0;i < n; ++i) dp[i][getchar() - 'a'] = 1;
for(int i = 0;i < m; ++i)
{
scanf("%d %d",&x,&y);
G[x - 1].push_back(y - 1);
in[y - 1]++;
}
int ans = -1;
if(TPS(n))
{;
memset(vis,false,sizeof(vis));
for(int i = 0;i < n; ++i)
{
if(!in[i])
{
vis[i] = true;
DFS(i);
for(int j = 0;j < 26; ++j) ans = max(ans,dp[i][j]);
}
}
}
printf("%d\n",ans);
return 0;
}